# Argue that no further mixed Nash Equilibria can exists

I'm looking at the following Normal-Form Game:

         B
|  1  |  2  |  3
---+-----------------
1| 1,-1|-1, 1|-1, 1
A 2|-1, 1| 2,-2|-1, 1
3|-1, 1|-1, 1| 3,-3


It's easy to see that there are no Nash-Equilibria in pure strategies. In mixed strategies I found, that $$(\sigma_A, \sigma_B)$$ is a Nash-Equilibrium with:

$$\sigma_A(1) = \sigma_B(1) = \frac{6}{13}$$ $$\sigma_A(2) = \sigma_B(2) = \frac{4}{13}$$ $$\sigma_A(3) = \sigma_B(3) = \frac{3}{13}$$

The problem sheet now asks to argue with an example why no further mixed Nash equilibrium can exist. I'm not sure how to argue here.

• What have you tried so far? Do you know what support sets are? Why are they forcing you to solve game theory exercises during summer? – Giskard Jul 30 '19 at 18:15
• @Giskard I never heard of support sets (I don't think they were included in my lecture) and I'm studying for an exam on Thursday. – user7802048 Jul 30 '19 at 18:39
• Here is the hint that @Giskard was trying to give you on support sets: "A pure strategy is in the support of a mixed strategy if that pure strategy is played with positive probability according to the mixed strategy. If a mixed strategy is played in a Nash equilibrium, all pure strategies in the support of that mixed strategy must yield an equal expected payoff." – Herr K. Jul 30 '19 at 19:17
• Also see my answer to a related question. – Herr K. Jul 30 '19 at 22:32